I have:
typedef struct
{
int id;
Other_Struct *ptr;
} My_Struct;
Lets say I have the pointer abcd which its type is My_Struct.
How can I get the address of:
abcd->ptr
?
I mean the address of ptr itself (like pointer to pointer) and not the address which is stored in ptr.
Just use the & address of operator, like this
Other_Struct **ptrptr = &(abcd->ptr);
How about this:
&(abcd->ptr)
If I understood correctly, in this scenario you have My_Struct *abcd pointing to an address, and what you want is the address of a field inside this structure (it doesn't matter if this field is a pointer or not). The field is abcd->ptr, so its address you want is &abcd->ptr.
You can easily check this by printing the actual pointer values (the difference between the addresses should give you the offset of ptr inside My_Struct):
struct My_Struct {
int id;
void *ptr;
};
main()
{
struct My_Struct *abcd;
printf("%p %p\n", abcd, &abcd->ptr);
}
Update: If you want your code to be portable, standards-compliant, and past and future proof, you may want to add casts to void * to the printf() arguments, as per @alk's comments below. For correctness, you can also use a standard entry point prototype (int main(void) or int main(int argc, char **argv)) and, of course, include stdio.h to use printf().
%p is defined for void-pointers only. So it shall be printf("%p %p\n", (void*) abcd, (void*) &abcd->ptr);
void(*)(), and usage such as the one above could be considered safe.)The most unambiguous way to do it is thus:
My_Struct *abcd;
Other_Struct **pptr;
pptr = &(abcd->ptr);
I don't know if the parentheses are really necessary, and I don't care, because it's more readable this way anyway.
